Problem: Let $V$ be a simple solid region oriented with outward normals that has a piecewise-smooth boundary surface $S$. $ \iiint_V \left[ \cos(x - y) + x - \sin(x + y) \right] \, dV$ Use the divergence theorem to rewrite the triple integral as a surface integral. Leave out constant coefficients and extraneous functions of $x$ and $z$. $ \oiint_S (e^{y} \hat{\imath} + $ $ \hat{\jmath} + (xz + y^2) \hat{k}) \cdot dS$
Explanation: Assume we have a simple solid region $V$ oriented with outward normals, and it has a piecewise-smooth, closed boundary surface $S$. If $F$ is a continuously differentiable vector field in $\mathbb{R}^3$, then the divergence theorem says: $ \oiint_S F \cdot dS = \iiint_V \text{div}(F) \, dV$ The given surface and boundary satisfy the conditions for the divergence theorem. We're converting a triple integral into a surface integral, so we know $\text{div}(F)$ and we want to find $F$. We already have two components: $F(x, y, z) = e^{y} \hat{\imath} + (???) \hat{\jmath} + (xz + y^2) \hat{k}$ Let's name the missing $y$ -component $P$. The only constraint we have on $P$ is that it must make $F$ have the correct divergence. $\begin{aligned} \text{div}(F) &= \dfrac{\partial P}{\partial y} + x \\ \\ &= \cos(x - y) + x - \sin(x + y) \end{aligned}$ Matching terms, our original constraint now becomes one specific requirement: $\dfrac{\partial P}{\partial y} = \cos(x - y) - \sin(x + y)$ The most general solution includes an arbitrary function of $x$ and $z$ that we can call $H(x, z)$. $P = \cos(x + y) - \sin(x - y) + H(x, z)$ The problem asks us to set $H(x, z) = 0$ so that we leave out extraneous functions of $x$ and $z$. Plugging $F$ into the divergence theorem, we conclude that an equivalent surface integral to the triple integral is: $ \oiint_S (e^y \hat{\imath} + (\cos(x + y) - \sin(x - y)) \hat{\jmath} + (xz + y^2) \hat{k}) \cdot dS$